# How do you solve using the completing the square method 2x^2 - 3x +1 = 0?

Apr 9, 2018

#### Explanation:

$2 {x}^{2} - 3 x + 1 = 0$
${x}^{2} - \frac{3 x}{2} + \frac{1}{2} = 0$
${x}^{2} - \frac{3 x}{2} + {\left(\frac{3}{4}\right)}^{2} = {\left(\frac{3}{4}\right)}^{2} - \frac{1}{2}$
${\left(x - \frac{3}{4}\right)}^{2} = \frac{9}{16} - \frac{8}{16}$
${\left(x - \frac{3}{4}\right)}^{2} = \frac{1}{16}$
$x - \frac{3}{4} = \pm \frac{1}{4}$
$x = \frac{3}{4} \pm \frac{1}{4}$

Therefore:
$x = 1 , \frac{1}{2}$